course Phy 121 ???L?~??????????Student Name:
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16:22:38 `q001. Note that this assignment contains 4 questions. If an object rotates through an angle of 20 degrees in five seconds, then at what rate is angle changing?
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RESPONSE --> We solve this problem as a velocity problem because we are talking about the angular velocity. angular v = `omega = `ds/`dt Therefore, the rate at which the angle is changing is 20 degrees divided by 5 seconds. `omega = 20deg/5s = 4 deg/s
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16:22:57 The change of 20 degrees in 5 seconds implies a rate of change of 20 degrees / (5 seconds) = 4 deg / sec. We call this the angular velocity of the object, and we designate angular velocity by the symbol `omega.
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RESPONSE --> I understand this concept.
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16:23:55 `q002. What is the average angular velocity of an object which rotates through an angle of 10 `pi radians in 2 seconds?
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RESPONSE --> We could convert radians to degrees but we don't have to. We can find angular velocity by dividing the angular displacement by the time interval. `omega = 10*`pi rad / 2sec = 5*`pi rad/sec
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16:24:08 The average angular velocity is equal to the angular displacement divided by the time required for that displacement, in this case giving us `omega = `d`theta / `dt = 10 `pi radians / 2 seconds = 5 `pi rad/s.
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RESPONSE --> I understand this problem.
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16:29:03 `q003. If an object begins with an angular velocity of 3 radians / sec and ends up 10 seconds later within angular velocity of 8 radians / sec, and if the angular velocity changes at a constant rate, then what is the average angular velocity of the object? In this case through how many radians this the object rotate and at what average rate does the angular velocity change?
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RESPONSE --> omega0 = 3rad/sec omegaF = 8rad/sec `dt = 10sec `d`omega = 5 rad/sec If we know the initial and final angular velocities, we can find the average. `omegaAve = (3rad/s + 8rad/s)/2 = 5.5 radians / second As with the the equations we have done in the past, displacement can be found by multiplying the time interval by the average angular velocity. `ds = `dt * `omegaAve = (10 s)(5.5rad/sec) = 55 radians Now, we can find angular acceleration by dividing the change in angular velocity by the time interval. alpha = `d`omega/`dt = (5 rad/s) / (10s) = 0.5 rad/s^2
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16:29:26 Starting at 3 rad/s and ending up at 8 rad/s, the average angular velocity would be expected to be greater than the minimum 3 rad/s and less than the maximum 8 rad/s. If the angular velocity changes at a constant rate, we would in fact expect the average angular velocity to lie halfway between 3 rad/s and 8 rad/s, at the average value (8 rad/s + 3 rad/s) / 2 = 5.5 rad/s. Moving at this average angular velocity for 10 sec the object would rotate through 5.5 rad/s * 10 s = 55 rad in 10 sec. The change in the angular velocity during this 10 seconds is (8 rad/s - 3 rad/s) = 5 rad/s; this change takes place in 10 seconds so that the average rate at which the angular velocity changes must be ( 5 rad / sec ) / (10 sec) = .5 rad/s^2. This is called the average angular acceleration. Angular acceleration is designated by the symbol `alpha. Since the angular velocity in this example changes at a constant rate, the angular acceleration is constant and we therefore say that `alpha = `d `omega / `dt. Again in this case `d`omega is the 5 rad/sec change in the angular velocity.
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RESPONSE --> I understand this problem.
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16:32:33 `q004. If an object starts out with angular velocity 14 rad/s and accelerates at a rate of 4 rad/s^2 for 5 seconds, then at what rate is the object rotating after the 5 seconds? Through how many radians will the object rotate during this time?
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RESPONSE --> omega0 = 14rad/s a = 4rad/s^2 `dt = 5s `d`omega = (a)(`dt) = (4rad/s^2)(5s) = 20 rad/s omegaF = 14rad/s + 20rad/s = 34rad/s at 5 seconds. `ds = `omegaAve * `dt = (14rad/s + 34rad/s)/2 * 5s = 120 radians
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16:32:52 Changing angular velocity at the rate of 4 rad/s^2 for 5 sec the angular velocity will change by (4 rad/s^2) (5s) = 20 rad/s. Since the angular velocity was already 14 rad/s at the beginning of this time period, it will be 14 rad/s + 20 rad/s = 34 rad/s at the end of the time period. The uniform rate of change of angular velocity implies that the average angular velocity is (14 rad/s + 34 rad/s) / 2 = 24 rad/s. An average angular velocity of 24 radians/second, in 5 seconds the object will rotate through an angle `d`theta = (24 rad/s) ( 5 sec) = 120 rad.
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RESPONSE --> I understand this problem.
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??????????? assignment #030 030. `query 30 Physics I 07-18-2006
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22:06:56 introductory set 8. If we know the constant moment of inertia of a rotating object and the constant net torque on the object, then how do we determine the angle through which it will rotate, starting from rest, in a given time interval?
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RESPONSE --> Torque is the product of the moment of intertia times the angular acceleration. This relationship is represented by the formula: tau = I * alpha. This relationship is much like the one between force and mass*acceleration. Therefore, if we know the moment of inertia and the net torque, we can rearrange the formula to find the angular acceleration. alpha = tau / I In order to find the angle through which the object rotates we use the relationship `ds = omegaAve * `dt However, we do not yet know the average angular velocity. We can find the change in angular velocity by multiplying the angular acceleration by the time interval. We know that the initial velocity is zero, so the change in velocity is equal to the final velocity. The average velocity would therefore be the final velocity divided by two.
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22:07:33 ** tau stands for torque and I stands for the moment of inertia. These quantities are analogous to force and mass. Just as F = m a, we have tau = I * alpha; i.e., torque = moment of inertia * angular acceleration. If we know the moment of inertia and the torque we can find the angular acceleration. If we multiply angular acceleration by time interval we get change in angular velocity. We add the change in angular velocity to the initial angular velocity to get the final angular velocity. In this case initial angular velocity is zero so final angular velocity is equal to the change in angular velocity. If we average initial velocity with final velocity then, if angular accel is constant, we get average angular velocity. In this case angular accel is constant and init vel is zero, so ave angular vel is half of final angular vel. When we multiply the average angular velocity by the time interval we get the angular displacement, i.e., the angle through which the object moves. **
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RESPONSE --> Ok.
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22:09:44 If we know the initial angular velocity of a rotating object, and if we know its angular velocity after a given time, then if we also know the net constant torque accelerating the object, how would we find its constant moment of inertia?
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RESPONSE --> We know, `omega0, `domega, `dt, tau and therefore we use the following formulas to solve for the constant moment of intertia. tau = I * alpha I = tau/alpha `d`omega = `omegaF - `omega0 alpha = `d`omega / `dt Now we just divide net torque by the average angular acceleration.
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22:10:06 ** From init and final angular vel you find change in angular vel (`d`omega = `omegaf - `omega0). You can from this and the given time interval find Angular accel = change in angular vel / change in clock time. Then from the known torque and angular acceleration we find moment of intertia. tau = I * alpha so I = tau / alpha. **
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RESPONSE --> I understand this problem.
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22:11:08 How do we find the moment of inertia of a concentric configuration of 3 uniform hoops, given the mass and radius of each?
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RESPONSE --> The formula of intertia for a hoop is the sum of the products of the mass and the radius squared. I = m1r1^2 + m2r2^2 + m3r3^2
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22:11:33 ** Moment of inertia of a hoop is M R^2. We would get a total of M1 R1^2 + M2 R2^2 + M3 R3^2. **
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RESPONSE --> Ok.
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22:13:39 How do we find the moment of inertia a light beam to which are attached 3 masses, each of known mass and lying at a known distance from the axis of rotation?
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RESPONSE --> The formula of intertia for a light beam is the same as the formula for a hoop. I = m1r1^2 + m2r2^2 + m3r3^2
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22:14:59 ** Moment of inertia of a mass r at distance r is m r^2. We would get a total of m1 r1^2 + m2 r2^2 + m3 r3^2. Note the similarity to the expression for the hoops. **
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RESPONSE --> Ok.
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22:17:06 Principles of Physics and General College Physics problem 8.4. Angular acceleration of blender blades slowing to rest from 6500 rmp in 3.0 seconds.
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RESPONSE --> `omegaF = 0 rpm `omega0 = 6500 rpm `dt = 3 sec `d`omega = `omegaF - `omega0 = 0rpm - 6500rpm = -6500 rpm Angular acceleration = alpha = `d`omega / `dt = -6500rpm / 3 sec = -2167 rpm/sec
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22:18:53 The change in angular velocity from 6500 rpm to rest is -6500 rpm. This change occurs in 3.0 sec, so the average rate of change of angular velocity with respect to clock time is ave rate = change in angular velocity / change in clock time = -6500 rpm / (3.0 sec) = -2200 rpm / sec. This reasoning should be very clear from the definition of average rate of change. Symbolically the angular velocity changes from omega_0 = 6500 rpm to omega_f = 0, so the change in velocity is `dOmega = omega_f - omega_0 = 0 - 6500 rpm = -6500 rpm. This change occurs in time interval `dt = 3.0 sec. The average rate of change of angular velocity with respect to clock time is therefore ave rate = change in angular vel / change in clock time = `dOmega / `dt = (omega_f - omega_0) / `dt = (0 - 6500 rpm) / (3 sec) = -2200 rpm / sec. The unit rpm / sec is a perfectly valid unit for rate of change of angular velocity, however it is not the standard unit. The standard unit for angular velocity is the radian / second, and to put the answer into standard units we must express the change in angular velocity in radians / second. Since 1 revolution corresponds to an angular displacement of 2 pi radians, and since 60 seconds = 1 minute, it follows that 1 rpm = 1 revolution / minute = 2 pi radians / 60 second = pi/30 rad / sec. Thus our conversion factor between rpm and rad/sec is (pi/30 rad / sec) / (rpm) and our 2200 rpm / sec becomes angular acceleration = 2200 rpm / sec * (pi/30 rad / sec) / rpm = (2200 pi / 30) rad / sec^2 = 73 pi rad / sec^2, or about 210 rad / sec^2.
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RESPONSE --> I did not convert my answer to radians per second^2. In order to do this, we multiply the quantity rpm/s^2 by (pi/30 rad/sec)rpm. alpha = 2167 rpm/sec * (pi/30 rad/sec)rpm = 72.23 pi*rad/sec^2
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22:23:26 Principles of Physics and General College Physics problem 8.16. Automobile engine slows from 4500 rpm to 1200 rpm in 2.5 sec. Assuming constant angular acceleration, what is the angular acceleration and how how many revolutions does the engine make in this time?
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RESPONSE --> `omega0 = 4500 rpm `omegaF = 1200 prm `dt = 2.5 sec `d`omega = 1200 rpm - 4500 rpm = -3300 rpm alpha = `d`omega/`dt = -3300rpm / 2.5sec = -1320 rpm/sec From the previous problem, we know how to convert these units to radians per second^2. alpha = (-1320 rpm/sec)(pi/30 rad/sec) / rpm = -44 pi*rad/s^2 `ds = `omegaAve * `dt `omegaAve = (1200rpm + 4500rpm)/2 = 2850 rpm `ds = 2850rpm * 2.5 sec = 7125 rpm*sec This means the object travels through 7125 revolutions during this time interval.
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22:24:09 The change in angular velocity is -3300 rpm, which occurs in 2.5 sec. So the angular acceleration is angular accel = rate of change of angular vel with respect to clock time = -3300 rpm / (2.5 sec) = 1300 rpm / sec. Converting to radians / sec this is about angular accel = -1300 rpm / sec ( pi / 30 rad/sec) / rpm = 43 pi rad/sec^2, approx.. Since angular acceleration is assumed constant, a graph of angular velocity vs. clock time will be linear so that the average angular velocity with be the average of the initial and final angular velocities: ave angular velocity = (4500 rpm + 1200 rpm) / 2 = 2750 rpm, so that the angular displacement is angular displacement = ave angular velocity * time interval = 2750 rpm * 2.5 sec = 6900 revolutions, approximately. In symbols, using the equations of uniformly accelerated motion, we could use the first equation `dTheta = (omega_0 + omega_f) / 2 * `dt = (4500 rpm + 1200 rpm) / 2 * (2.5 sec) = 6900 revolutions and the second equation omega_f = omega_0 + alpha * `dt, which is solved for alpha to get alpha = (omega_f - omega_0) / `dt = (4500 rpm - 1200 rpm) / (2.5 sec) = 1300 rpm / sec, which as before can be converted to about 43 pi rad/sec^2, or about 130 rad/sec^2. The angular displacement of 6900 revolutions can also be expressed in radians as 6900 rev = 6900 rev (2 pi rad / rev) = 13800 pi rad, or about 42,000 radians.
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RESPONSE --> I did not convert the revolutions to be expressed in radians. In order to do this, we multiply the number of revolution by 2 pi*rad/rev (7125 rpm/s)(2 pi*rad/rev) = 14250 pi*rad
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22:28:58 gen Problem 8.23: A 55 N force is applied to the side furthest from the hinges, on a door 74 cm wide. The force is applied at an angle of 45 degrees from the face of the door. Give your solution:
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RESPONSE --> F = 55 N `ds = width = 74 cm = 0.74 m angle = 45 deg The torque perpendicular to the door is the perpendicular force multiplied by the moment arm which I have represented as `ds. tau = F * `ds tau perpendicular = 55N * 0.74m = 40.7 N*m The magnitude of the torque at this angle is in the y direction. Therefore, to find the torque we use the following equation: tau = F perp * sin(45deg) = (40.7 N*m)(sin 45 deg) = 28.8 N*m
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22:31:22 ** ** If the force is exerted perpendicular to the face of the door, then the torque on the door is 55 N * .74 m = 40.7 m N. The rest of the given solution here is for a force applied at an angle of 60 degrees. You can easily adapt it to the question in the current edition, where the angle is 45 degrees:The torque on the door is 45 N * .84 m = 37.8 m N. If the force is at 60 deg to the face of the door then since the moment arm is along the fact of the door, the force component perpendicular to the moment arm is Fperp = 37.8 m N * sin(60 deg) = 32.7 N and the torque is torque = Fperp * moment arm = 32.7 N * .84 m = 27.5 m N. STUDENT COMMENT: Looks like I should have used the sin of the angle instead of the cosine. I was a little confused at which one to use. I had trouble visualizing the x and y coordinates in this situation.INSTRUCTOR RESPONSE: You can let either axis correspond to the plane of the door, but since the given angle is with the door and angles are measured from the x axis the natural choice would be to let the x axis be in the plane of the door. The force is therefore at 60 degrees to the x axis. We want the force component perpendicular to the door. The y direction is perpendicular to the door. So we use the sine of the 60 degree angle.
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RESPONSE --> My answers agreed with the solutions in the book. Was my thinking process correct? There were some numbers in this solution that I'm not sure where they came from.
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22:35:34 gen problem 8.11 rpm of centrifuge if a particle 7 cm from the axis of rotation experiences 100,000 g's
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RESPONSE --> I was not required to complete this text problem.
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22:36:36 ** alpha = v^2 / r so v = `sqrt( alpha * r ) = `sqrt( 100,000 * 9.8 m/s^2 * .07 m) = `sqrt( 69,000 m^2 / s^2 ) = 260 m/s approx. Circumference of the circle is 2 `pi r = 2 `pi * .07 m = .43 m. 260 m/s / ( .43 m / rev) = 600 rev / sec. 600 rev / sec * ( 60 sec / min) = 36000 rev / min or 36000 rpm. All calculations are approximate. **
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RESPONSE -->
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22:40:53 gen problem 8.20 small wheel rad 2 cm in contact with 25 cm wheel, no slipping, small wheel accel at 7.2 rad/s^2. What is the angular acceleration of the larger wheel?
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RESPONSE --> r1 = 2cm r2 = 25cm a1 = 7.2 rad/s^2 Since r1 is 2/25 of r2, and radius is directly related to velocity and acceleration, we can assume that a1 is 2/25 of a2. Therefore, to find a2 we multiply a1 by 2/25. a2 = 2/25 * 7.2rad/s^2 = 0.576 rad/s^2
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22:41:41 ** Since both wheels travel the same distances at the rim, angular displacements (which are equal to distance along the rim divided by radii) will be in inverse proportion to the radii. It follows that angular velocities and angular accelerations will also be in inverse proportion to radii. The angular acceleration of the second wheel will therefore be 2/25 that of the first, or 2/25 * 7.2 rad/s^2 = .58 rad/s^2 approx.. **
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RESPONSE --> I understand this problem.
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22:43:46 How long does it take the larger wheel to reach 65 rpm?
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RESPONSE --> First we must have the same units involved in all the variables. Convert rpm's to rad/s by multiplying by pi/30 rad/sec. (65 rpm)(pi/30 rad/sec) = 2.17 pi*rad/sec. This is the change in angular velocity. If the larger wheel's angular acceleration is 0.576 rad/s^2, we can find the time interval over a certain distance this way: `dt = `d`omega / `alpha = (2.17 pi*rad/sec) / (0.576 rad/s^2) = 11.8 seconds
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22:43:51 ** 65 rpm is 65 * 2 `pi rad / min = 65 * 2 `pi rad / (60 sec) = 6.8 rad / sec, approx. At about .6 rad/s/s we get `dt = (change in ang vel) / (ang accel) = 6.8 rad / s / ( .6 rad / s^2) = 11 sec or so. **
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RESPONSE --> I understand this problem.
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22:44:11 Univ. 9.72 (64 in 10th edition). motor 3450 rpm, saw shaft 1/2 diam of motor shaft, blade diam .208 m, block shot off at speed of rim. How fast and what is centrip accel of pt on rim?
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RESPONSE --> This problem is above my course level.
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22:44:14 ** The angular velocity of the shaft driving the blade is double that of the motor, or 3450 rpm * 2 = 7900 rpm. Angular velocity is 7900 rpm = 7900 * 2 pi rad / 60 sec = 230 pi rad / sec. At a distance of .208 m from the axis of rotation the velocity will be .208 m * 230 pi rad / sec = 150 m/s, approx.. The angular acceleration at the .208 m distance is aCent = v^2 / r = (150 m/s)^2 / (.208 m) = 108,000 m/s^2, approx.. The electrostatic force of attraction between sawdust and blade is nowhere near sufficient to provide this much acceleration. **
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RESPONSE -->
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